Answer
$\displaystyle \frac{\sec^{4}x}{4}+C$
Work Step by Step
$\displaystyle \sin x\sec^{5}x=\frac{\sin x}{\cos^{5}x}=\tan x\sec^{4}x$
$I=\displaystyle \int\sin x\sec^{5}xdx=\int\sec^{2}x(\sec^{2}x\tan x)dx$
Substitute: $\left[\begin{array}{l}
u=\sec^{2}x\\
du=2\sec^{2}x\tan x
\end{array}\right],\quad$
$I=\displaystyle \int\frac{u}{2}du$
$=\displaystyle \frac{1}{2}\cdot\frac{u^{2}}{2}+C$
bring back x
$=\displaystyle \frac{\sec^{4}x}{4}+C$
